PHYSICS: E. DERSHEM 
63 
Some of the conditions affecting the resolving power of an X-ray spectrom- 
eter, that is the ability of the instrument to separate two waves of nearly 
the same length may be derived by the aid of the diagram (fig. 1). 
Assume a source, i.e., a slit, of width s and a crystal of thickness / and 
assume that the absorption in the crystal is not so great but that some of 
the rays may penetrate entirely through the crystal and being reflected from 
the planes on the lower side again traverse the crystal and finally reach the 
photographic plate B'A'C'D'. It is easily seen that there will be an image 
on the plate equal in width to the width of the source s, due to reflection 
from the upper surface alone. In addition there is a widening of the line 
due to the part reflected from the lower planes equal to the line DE which is 
drawn from P perpendicular to AA f . Then since DF = t = the thickness of 
the crystal, t = AD sin 6 and AD = DE/sin 2 0 and by substitution 
t = DE/sin 0 = \ DE/cos 0. Whence DE = 2 t cos 0. 
Since DE is the width of beam due to penetration into the crystal the total 
width of beam is s + 2 t cos 0 and this is the width of line on the photo- 
graphic plate. 
Crystal 
FIG. 1 FIG. 2 
In order to resolve two lines of nearly the same wave length it is neces- 
sary that their images on the photographic plate should not overlap, that is 
the centers of their images must be further apart than the width of beam, 
s + 2 t cos 0. 
Assume two wave lengths, X and X + AX, then to find how small AX may 
be and still have the two wave lengths clearly separated on the plate: Using 
the formula n\ = 2 d sin 0 let X take on a small increment AX and 0 the 
corresponding increment Ad. Then by differentiation we obtain nA\ = 
d cos 6 Ad. 
From figure 2 we see that the angle of the crystal must be changed by the 
amount A0 in order to reflect the wave of length X + AX instead of the one 
of length X and that the reflected ray being rotated through twice this amount 
is rotated through the angle 2A0. If the distance from the crystal to the 
plate is r then the distance the beam has moved along the plate in changing 
from X to X + AX is 2rAd and this distance must be greater than the width of 
beam, s -\- 2 t cos 6. Thus 
2rAS > s + 2t cos 0 
